JEE Advanced · Mathematics · 12. Circle
Let the straight line \(\mathrm{y}=2 \mathrm{x}\) touch a circle with center \((0, \alpha), \alpha>0\), and radius \(\mathrm{r}\) at a point \(\mathrm{A}_1\). Let \(\mathrm{B}_1\) be the point on the circle such that the line segment \(A_1 B_1\) is a diameter of the circle. Let \(\alpha+r=5+\sqrt{5}\).
Match each entry in List-I to the correct entry in List-II.
The correct option is
- A \((\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(3)\)
- B \((\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(3)\)
- C \((\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(5) \quad(\mathrm{S}) \rightarrow(3)\)
- D \((\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(3) \quad(\mathrm{S}) \rightarrow(5)\)
Answer & Solution
Correct Answer
(C) \((\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(5) \quad(\mathrm{S}) \rightarrow(3)\)
Step-by-step Solution
Detailed explanation
Consider centre as \(\mathrm{P}(0, \alpha), \alpha>0\)
\(\left|\frac{2(0)-\alpha}{\sqrt{5}}\right|=r\)
\(\begin{aligned}& |-\alpha|=\sqrt{5} r \\& \alpha=\sqrt{5} r \\& \therefore \alpha+r=5+\sqrt{5} \\& \sqrt{5} r+r=\sqrt{5}(\sqrt{5}+1) \\& r=\sqrt{5}, \alpha=5 \\& \therefore P(0,5)\end{aligned}\)
Foot of perpendicular from \(\mathrm{P}\) to line \(2 \mathrm{x}-\mathrm{y}=0\)
\(\begin{aligned}& \frac{x-0}{2}=\frac{y-5}{-1}=\frac{-(2(0)-5)}{5}=1 \\& x=2, y=4 \quad A_1(2,4)\end{aligned}\)
Let \(\mathrm{B}(\mathrm{p}, \mathrm{q})\)
\(\therefore \frac{\mathrm{p}+2}{2}=0, \frac{\mathrm{q}+4}{2}=5\)
\(\therefore \mathrm{p}=-2, \mathrm{q}=6 \quad \mathrm{~B}(-2,6)\)
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